Frequency Domain Representation of Signals

Transcription

1 Frequency Domain Representation of Signals The Discrete Fourier Transform (DFT) of a sampled time domain waveform x n x 0, x 1,..., x 1 is a set of Fourier Coefficients whose samples are 1 n0 X k X0, X 1,..., X 1 where Xk xn * e nk j The inverse DFT converts the Fourier Coefficients to the corresponding time domain samples where 1 1 j x[ n] X[ k] e k0 The Discrete Fourier Transform (DFT) allows us to investigate discrete time signals in the frequency domain. The DFT is used to represent a discrete time signal as the superposition of sine waves. Fast Fourier Transform (FFT) is an algorithm to compute the DFT efficiently when the number of points is a power of, i.e., = k where k is an integer. X k X k jx k x n, X k X * k. X k R I nk is complex, and for Real X R k X I k is called the Magnitude or Amplitude Spectrum. For Real x[n], the Magnitude Spectrum is symmetric about k=/. The Phase Spectrum is: Phase X k 1 X tan X I k k R * The Power Spectrum, Sk Xk X k Xk Page 1 of 11

2 DFTs and ADCs If x[n] are samples from an Analog to Digital Converter (ADC) operating at F s x[n] is Real. k * Fs X[k] represents frequency components at Frequency k. 1 Fs f is the spacing of the points or bins on the x-axis. f. Therefore, Ts the frequency resolution is inversely proportional to the total sampling time (T s ). Because the Magnitude Spectrum is symmetric about /, we usually only plot X[k] for k < /. The maximum frequency represented is Fs/. Units of Power Spectrum is Watts because we assume that the signal is in volts delivered to a 1 ohm resistor so x[n] = V[n] = I[n] therefore P[n] = x[n] Watts. S The Power Spectral Density, k PSD k and has units Watts/Hz. f Measure Power by integrating the Power Spectral Density. Take square root to obtain V rms. That is Pwr = V rms ote that if two signals have power P 1 and P and associated V rms1 and V rms then P 1 + P = P Total and V rmstotal V rms1 + V rms. V rmstotal = sqrt (P Total ). Page of 11

3 Example: 10 periods of 1 Volt, 50KHz sine wave, sampled at 1MHz (00 points) ote that we have 0 samples per cycle and exactly 10 cycles so no window is needed (see Windowing section below). 1 xn Asin f nt where A=1, f = 50KHz, and T 1 us Fs Quiz: V Peak = V V RMS = V = dbv rms Power 1ohm Resistor = W = dbw Power Spectral Density ( f =5KHz) at 50KHz = W/Hz Bandwidth of sampled signal is 500KHz. 500KHz 1MHz are the image frequencies. There are 00 points in the Double Sided DFT. Amplitude in the Double Sided DFT in the 50KHz and the 950KHz bin is 0.5 V in both peaks. Amplitude in the Single Sided DFT in the 50KHz bin is 1.0 V Bandwidth = 500KHz. Page 3 of 11

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5 (Zooming in around peak) A 1 Peak is at 0log( ) 0log( ) -3dBVrms A 1 Peak is at 10log( ) 10log( ) -3dBW What is setting the oise Floor here? (Zooming in around peak) Page 5 of 11

6 Peak is at A f 1 1MHz 00 W Hz Page 6 of 11

7 Pulse Train Pulse train has same shape in Time Domain and Frequency Domain. Square Wave Square Wave only has odd harmonics at with amplitudes 1/3, 1/5, 1/7, of the fundamental. 1/3 = -9.5 db, 1/5 = -14 db Page 7 of 11

8 White Gaussian oise (WG) Standard Deviation Vrms 99.7% of values within 3 for White Gaussian oise (WG) PSD is flat for WG Variance oise Power PSD df Fs For WG, PSDdf 0 * BW 0 BW For this example nw 0 Average of PSD 500, Fs 1MHz Hz 9 1e6 oisepower 500*10 * 0.5W BW Vrms V From Histogram or Time Domain Graph, Maximum Peak-Peak 1.5V 3 Page 8 of 11

9 Windowing The DFT assumes that the samples are periodic. In the above examples, there were exactly 10 periods of the sine wave. However, what if there are 10.5 periods. Windowing Example, (Installation Instructions) This discontinuity causes Spectral Leakage into all frquencies. We aren t getting the 350dB attenuation outside the signal bins anymore (see Frequency Spectrum for 10.0 periods). This causes errors at all frequencies including the signal bins. The peak is now at 4dBVrms instead of 3. Spectral leakage can be virtually eliminated by windowing time domain samples prior to calculating the DFT. o Windows taper smoothly down to zero at the beginning and the end of the observation window (eliminating any discontinuities). o Time domain samples are multiplied by the window on a sample by sample basis. o Multiplication in the Time Domain is Convolution in the Frequency Domain. o The effect of the Convolution is to smear our peaks in the Frequency domain. We will use the Hodie Window in this class. This window smears the peaks over 17 bins in the Frequency Domain. Page 9 of 11

10 (Zooming in around peak) To measure the amplitude after windowing, you have to integrate the Power Spectral Density over these 17 Signal Bins. Page 10 of 11

11 Once again we are back to the Power = 0.5W = -3dBW, or A=1.0 Why 10KHz to 90KHz? References: Mixed-Signal System Design and Modeling, Eric Swanson, Lecture 13, Fall Semester 00. Page 11 of 11